All categories

2 years ago

(−m 2 +6)+(−4m 2 +7m+2)

Mathematics

1 answer:

dmitriy555 [2]2 years ago

7 0

Following are calculation to the given eqution:

Given:

$\bold{(-m^2+6)+(-4m^2+7m+2)}$

To find:

Solve the equation.

Solution:

$\to \bold{(-m^2+6)+(-4m^2+7m+2)}\\\\\to \bold{-m^2+6-4m^2+7m+2}\\\\\to \bold{-5m^2+7m+8}\\\\\to \bold{-1(5m^2-7m-8)}\\\\$

Therefore, the final solution is " $-1(5 m^2-7m-8)$ ".

Learn more:

brainly.com/question/12685776

You might be interested in

Pleaser answer CORRECTLY , WILL MARK BRAINLIEST !!!!!!!!!

Nataly [62]

Answer:

Step-by-step explanation:

-3a + 6b = a + 4b

-4a + 6b = 4b

-4a = -2b

a = -2b/-4

f(a) = -b/2

5 0

3 years ago

Select the symbol that represents the phrase "at most."

anastassius [24]

Your answer will be c

4 0

3 years ago

Consider the function below. f(x)= x^3 + 2x^2 - x - 2 plot the x and y intercepts of the function

dimulka [17.4K]

In the Figure below is shown the graph of this function. We have the following function:

$f(x)=x^3+2x^2-x-2$

The $y-intercept$ occurs when $x=0$ , so:

$f(0)=(0)^3+2(0)^2-(0)-2=-2$

Therefore, the $y-intercept$ is the given by the point:

$\boxed{(0,-2)}$

From the figure we have three $x-intercepts$ :

$\boxed{P_{1}(-2,0)} \\ \boxed{P_{2}(-1,0)} \\ \boxed{P_{3}(1,0)}$

So, the $x-intercepts$ occur when $y=0$ . Thus, proving this:

$f(x)=x^3+2x^2-x-2 \\ \\ For \ P_{1}:\\ If \ x=-2, \ y=(-2)^3+2(-2)^2-(-2)-2=0 \\ \\ For \ P_{2}:\\ If \ x=-1, \ y=(-1)^3+2(-1)^2-(-1)-2=0 \\ \\ For \ P_{3}:\\ If \ x=1, \ y=(1)^3+2(1)^2-(1)-2=0$

7 0

3 years ago

Read 2 more answers

Need help asap will give brainliest

ch4aika [34]

horizontal shift: 3 units to the left
vertical shift: 4 units down

4 0

2 years ago

Read 2 more answers

Evaluate -3- (-4) - (-2) + 1.

Shtirlitz [24]

Answer:

Step-by-step explanation:

-3+4+2+1=4

8 0

2 years ago